solution of non linear differential second order equation

Question

we have the following differential equation $$ uu''=1+(u') ^2 $$ i found that the general solution of this equation is $$ u=d \cosh ((x-b)/d) $$ where $b$ and $d$ are constats.

Please how we found this general solution?

Answer 1

You can separate that equation as $$ \frac{2u'u''}{1+u'^2}=2\frac{u'}{u} $$ where both sides are complete differentials which integrate to $$ \ln(1+u'^2)=\ln(u^2)+c\implies 1+u'^2=Cu^2. $$ Can you continue?

Answer 2

$$uu''=1+(u') ^2$$ Substitute $p=u'$ $$u\frac {dp}{dx}=1+p^2$$ $$u\frac {dp}{du}\frac {du}{dx}=1+p^2$$ $$u\frac {dp}{du}p=1+p^2$$ Now it's separable $$\int \frac {pdp}{1+p^2}=\int \frac{du}u$$ It should be easy to integrate now..

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Edit $$p^2+1=Ku^2 \implies \int \frac {du}{\sqrt {Ku^2-1}}=\pm x+K_2$$ $$\frac {arcosh(\sqrt K u)}{\sqrt K}=x+K_2$$ Taking $cosh$ on both side $$\sqrt K u=\cosh({\sqrt K}(x+K_2))$$ $$ \boxed{u=\frac 1 {\sqrt K}\cosh(\sqrt K x+K_2)}$$ Which is close to your formula $$u=d \cosh ((x-b)/d) \implies d=1/\sqrt K \text{ and } -b/d=K_2$$